1005、K次取反后最大化的数组和

讲解：https://programmercarl.com/1005.K%E6%AC%A1%E5%8F%96%E5%8F%8D%E5%90%8E%E6%9C%80%E5%A4%A7%E5%8C%96%E7%9A%84%E6%95%B0%E7%BB%84%E5%92%8C.html

关键：绝对值排序

class Solution {
public:
    static bool cmp(int a, int b) {
    return abs(a) > abs(b);
}
    int largestSumAfterKNegations(vector<int>& nums, int k) {
        sort(nums.begin(), nums.end(), cmp);
        for (int i=0; i< nums.size();i++){
            if(nums[i] < 0 && k > 0) {
                nums[i] *= -1;
                k--;
            }
        }
        if (k % 2 == 1) nums[nums.size() -1 ] *= -1;
        int result =0;
        for (int a:nums) result+=a;
        return result;
    }
};

134、加油站

讲解：https://programmercarl.com/0134.%E5%8A%A0%E6%B2%B9%E7%AB%99.html

class Solution {
public:
    int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
        int curSum = 0;
        int totalSum = 0;
        int start = 0;
        for (int i = 0; i<gas.size(); i++){
            curSum += gas[i] - cost[i];
            totalSum += gas[i] - cost[i];
            if (curSum < 0){
                start = i+1;
                curSum = 0;
            }
            
        }
        if (totalSum < 0) return -1;
        return start;
    }
};

135、分发糖果

讲解：https://programmercarl.com/0135.%E5%88%86%E5%8F%91%E7%B3%96%E6%9E%9C.html

从前到后 和 从后到钱 两遍贪心遍历。

class Solution {
public:
    int candy(vector<int>& ratings) {
        vector<int> candyVec(ratings.size(), 1);
        for (int i=1; i < ratings.size(); i++){
            if (ratings[i-1] < ratings[i]) candyVec[i] = candyVec[i-1] +1;
        }
        for (int i=ratings.size()-1; i>0;i--){
            if (ratings[i-1]>ratings[i]) candyVec[i-1] = max(candyVec[i-1], candyVec[i]+1);
        }
        int result = 0;
        for (int r:candyVec) result+=r;
        return result;
    }
};